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LD
23 tháng 9 2017
a) \(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)ư
=\(\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x +2\sqrt{x}-7-x+\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4}\)
=\(\dfrac{-4\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
b)ta có : \(\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+2}{\sqrt{x}-3}=1+\dfrac{2}{\sqrt{x}-3}\)
để P nguyên thì \(\sqrt{x}-3\inƯ\left(2\right)\Leftrightarrow\sqrt{x}-3\inƯ\left(\pm1,\pm2\right)\)
\(\Rightarrow\sqrt{x}-3=1\Leftrightarrow x=16\left(TM\right)\)
\(\sqrt{x}-3=-1\Leftrightarrow x=4\left(KTM\right)\)
\(\sqrt{x}-3=2\Leftrightarrow x=25\left(TM\right)\)
\(\sqrt{x}-3=-2\Leftrightarrow x=1\left(KTM\right)\)
vậy x\(\in\left\{16,25\right\}\)
16 tháng 6 2017
a) P = \(\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
= \(\left(\dfrac{-\sqrt{x}+\sqrt{x}+1}{\sqrt{x}+1}\right)\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\left(\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\left(\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{x-9-\left(x-4\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\left(\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\left(\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\dfrac{1}{\sqrt{x}+1}:\dfrac{1}{\sqrt{x}-2}\) = \(\dfrac{1}{\sqrt{x}+1}.\dfrac{\sqrt{x}-2}{1}\) = \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
NC
18 tháng 8 2020
a) \(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\) \(\left(x\ge0;x\ne1\right)\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)
NC
18 tháng 8 2020
b) \(P=\frac{7}{2}\)
\(\Leftrightarrow\frac{3\sqrt{x}+8}{\sqrt{x}+2}=\frac{7}{2}\)
\(\Rightarrow6\sqrt{x}+16=7\sqrt{x}+14\)
\(\Leftrightarrow\sqrt{x}=2\Rightarrow x=4\)
27 tháng 8 2019
a, ĐKXĐ : \(x> 0 ; x \neq 1 \)
P = \(\dfrac{3x+3\sqrt{x} - 3}{\sqrt{x^2} +2\sqrt{x} - \sqrt{x} - 2}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{\sqrt{x}} . \dfrac{1-( 1 -\sqrt{x})}{1-\sqrt{x}}\)
= \(\dfrac{3x+3\sqrt{x} - 3 }{\sqrt{x}(\sqrt{x}+2)-(\sqrt{x} - 2)}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{\sqrt{x}}. \dfrac{ 1-1+\sqrt{x}}{1-\sqrt{x}}\)
= \(\dfrac{3x+3\sqrt{x} - 3 }{(\sqrt{x}+2)(\sqrt{x}-1)}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{(\sqrt{x}-1)} \)
= \(\dfrac{3x+3\sqrt{x}-3-(\sqrt{x}-1)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
= \(\dfrac{3x+3\sqrt{x}-3-(\sqrt{x^2}- 1^2) - (\sqrt {x^2}-2^2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
= \(\dfrac{3x+3\sqrt{x} - 3 - x+1-x+4}{(\sqrt{x}+2)(\sqrt{x}-1)} \)
= \(\dfrac{x+3\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x} - 1)}\)
= \(\dfrac{\sqrt{x^2}+2\sqrt{x} +\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x} - 1)} \)
= \(\dfrac{\sqrt{x}(\sqrt{x}+2)+(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x} - 1)} \)
= \(\dfrac{(\sqrt{x}+2)(\sqrt{x}+1)}{(\sqrt{x}+2)(\sqrt{x} - 1)} \)
= \(\dfrac{\sqrt{x}+1}{\sqrt{x} - 1} \)
c, Để P = \(\sqrt{x}\) \(\Leftrightarrow\) \(\dfrac{\sqrt{x}+1}{\sqrt{x} - 1} \) = \(\sqrt{x} \)
\(\Rightarrow\) \(\sqrt{x}+1= \sqrt{x}(\sqrt{x}-1)\)
\(\Leftrightarrow\) \(\sqrt{x}+1 = \sqrt{x^2} - \sqrt{x}\)
\(\Leftrightarrow\) \( \sqrt{x^2} -\sqrt{x} - \sqrt{x} - 1 = 0\)
\(\Leftrightarrow\) \(\sqrt{x^2} - 2\sqrt{x} +1-1-1=0\)
\(\Leftrightarrow\) \((\sqrt{x}-1)^2 - (\sqrt{2})^2 \) = 0
\(\Leftrightarrow\) \((\sqrt{x} - 1 - \sqrt{2})(\sqrt{x} - 1+\sqrt{2})\)
\(\Leftrightarrow\) \(\begin{cases} \sqrt{x} - 1 - \sqrt{2}=0 \\ \sqrt{x} - 1 +\sqrt{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} \sqrt{x} = 1 +\sqrt{2} \\ \sqrt{x} = 1 - \sqrt{2} \end{cases} \) \(\Leftrightarrow\)\(\begin{cases} x = 1+\sqrt{2} = 3+2\sqrt{2} \\ \sqrt{x} = 1-\sqrt{2} < 0 ( LOẠI ) \end{cases} \)
P/s : mk không biết làm phần b
26 tháng 2 2020
M = \(\frac{2\sqrt{x}-9x}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{9-x+2x-3\sqrt{x}}{x-5\sqrt{x}+6}\)
=\(\frac{x-\sqrt{x}}{x-5\sqrt{x}+6}\)